# Any mechanics ppl out there willing to help me on this question (exercise 3d of m1)

Watch
Announcements
This discussion is closed.
#1
i dne questions 2-8 (except part b of 8) i need real help on this question!

A ball is thrown vertically upwards with a speed of 10ms from a point 2m above the ground

a) it wants the length of time for which the ball is 3m or more above the ground
b)and the speed at which it hits the ground

i tried it like 10 times and i got like slightly different to the back of the book. could ne1 tell me how to do this question, i actually dnt the other 'particle 1's' except this 1 HELP (
0
16 years ago
#2
(Original post by Godmaster)
i dne questions 2-8 (except part b of 8) i need real help on this question!

A ball is thrown vertically upwards with a speed of 10ms from a point 2m above the ground

a) it wants the length of time for which the ball is 3m or more above the ground
b)and the speed at which it hits the ground

i tried it like 10 times and i got like slightly different to the back of the book. could ne1 tell me how to do this question, i actually dnt the other 'particle 1's' except this 1 HELP (
a)s=ut+1/2 at^2 use that equationa nd substitute...

1 = 10t - 4.9t^2....use quadratic equation....should get 2 times....minus smaller one from bigger one.....

b) use.....v^2=u^2+2as

v^2 = 100 - 19.6 (-2)
and solve for v
0
16 years ago
#3
(Original post by Godmaster)
i dne questions 2-8 (except part b of 8) i need real help on this question!

A ball is thrown vertically upwards with a speed of 10ms from a point 2m above the ground

a) it wants the length of time for which the ball is 3m or more above the ground
b)and the speed at which it hits the ground

i tried it like 10 times and i got like slightly different to the back of the book. could ne1 tell me how to do this question, i actually dnt the other 'particle 1's' except this 1 HELP (
Work out the total time for the flight of the ball then subtract the time when the ball is below 3m high (this can be found by forming a quadratic using x=ut+1/2at^2). The speed at which the ball lands can be found by finding the height the ball reaches, then calculating the speed it will have after accelarating downwards due to gravity(v^2=u^2 + 2ax).
0
#4
(Original post by matouwah)
a)s=ut+1/2 at^2 use that equationa nd substitute...

1 = 10t - 4.9t^2....use quadratic equation....should get 2 times....minus smaller one from bigger one.....

b) use.....v^2=u^2+2as

v^2 = 100 - 19.6 (-2)
and solve for v
1st 1 i tried but it wants above 3m not 2 grr annoying. 2nd 1 i get 8.97ms but that isnt wot it ses at the bk of the book lol.

thnx for trying 0
16 years ago
#5
(Original post by Godmaster)
1st 1 i tried but it wants above 3m not 2 grr annoying. 2nd 1 i get 8.97ms but that isnt wot it ses at the bk of the book lol.

thnx for trying well if it wants above 3m....the displacement will be 1m as u start 2m above ground...so i'm not sure why it doesnt work.......
0
16 years ago
#6
(Original post by Godmaster)
1st 1 i tried but it wants above 3m not 2 grr annoying. 2nd 1 i get 8.97ms but that isnt wot it ses at the bk of the book lol.

thnx for trying Well work it out for above 2m. You can work out how long it is bloew 3m for, by saying:

u=10
x=1 (3-2)
a=-9.81

x=ut+1/2at^2
9.81/2t^2 - 10t + 1 =0

t= (10+/-(100-2(9.81))^1/2)9.81
........
0
16 years ago
#7
No no no don't do that for (b)

use original data, displacement is 2 m to ground so u=-10, v=v, s=3, a=g and t='who cares'

v^2=u^2 + 2as
v^2=100 +4g
v=sqrt(100 + 4g)
v=11.7983...
v=11.8m/s
0
16 years ago
#8
Before you ask, set down as the positive direction so acceleration and displacement are +ve values and the initial velocity, upwards becomes negative.

Wherever possible always use data in question to answer supplimentary parts to a problem. The methods descibed above using maximum heights, time above 3m etc are just inviting careless mistakes and round off errors in subsequent calcs.
0
16 years ago
#9
(Original post by Godmaster)
i dne questions 2-8 (except part b of 8) i need real help on this question!

A ball is thrown vertically upwards with a speed of 10ms from a point 2m above the ground

a) it wants the length of time for which the ball is 3m or more above the ground
b)and the speed at which it hits the ground

i tried it like 10 times and i got like slightly different to the back of the book. could ne1 tell me how to do this question, i actually dnt the other 'particle 1's' except this 1 HELP (
are these right?

a) = 1.83s
b) = -11.80ms^-1
0
X
new posts Back
to top

view all
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes, my partner and I are struggling (13)
6.91%
Yes, my partner and I broke up (14)
7.45%
Yes, it's hard being around my family so much (43)
22.87%
Yes, I'm feeling lonely isolating alone (24)
12.77%
No, nothing has changed (64)
34.04%
No, it's helped improve my relationships (30)
15.96%